# How do you evaluate  e^( ( pi)/4 i) - e^( ( 11 pi)/6 i) using trigonometric functions?

May 26, 2017

${e}^{\frac{\pi}{4} i} - {e}^{\frac{11 \pi}{6} i} = \frac{\sqrt{2} - \sqrt{3}}{2} + \frac{\sqrt{2} + 1}{2} i$

#### Explanation:

There is a special formula for complex numbers that we will use to solve this problem:

${e}^{i \theta} = \cos \theta + i \sin \theta$

Therefore, we can rewrite this problem as:

${e}^{\frac{\pi}{4} i} - {e}^{\frac{11 \pi}{6} i}$

$= \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right) - \cos \left(\frac{11 \pi}{6}\right) - i \sin \left(\frac{11 \pi}{6}\right)$

Now all we have to do is simplify.

$= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i - \frac{\sqrt{3}}{2} - \left(- \frac{1}{2} i\right)$

$= \frac{\sqrt{2} - \sqrt{3}}{2} + \frac{\sqrt{2} + 1}{2} i$

So ${e}^{\frac{\pi}{4} i} - {e}^{\frac{11 \pi}{6} i} = \frac{\sqrt{2} - \sqrt{3}}{2} + \frac{\sqrt{2} + 1}{2} i$